Question

(Definite intergral) (1/ x^2 (radx^2 +1))dx using trig functions

Answer 1

what does the (x^2+1) part tell us about which sub we use?

Answer 2

\[\int\limits_{}^{}\frac{1}{x^2 \sqrt{x^2+1}} dx?\]

Answer 3

that we can let u = tan u

Answer 4

Yes

Answer 5

close. let x=tan(u)

Answer 6

ok, my apologies

Answer 7

sorry doesn't cut it i was offended when u said let u=tan u
lol

Answer 8

lol

Answer 9

ok so what is the next step?

Answer 10

This is my first time doing a problem like this

Answer 11

take derivative of both sides

Answer 12

my comp froze :(

Answer 13

i believe we differenate

Answer 14

right! :)

Answer 15

ok

Answer 16

is this right
if x=tan(u)
then
dx=sec^2(u) du
?

Answer 17

I agree myininaya

Answer 18

ok now lets make our substitutions in the integral

Answer 19

\[\int\limits_{}^{}\frac{1}{(\tan(u))^2 \sqrt{tan(u))^2+1}} \sec^2(u) du\]

Answer 20

recall your trig identity \[\tan^2(u)+1=\sec^2(u)\]

Answer 21

ok I do and I got that part :-)

Answer 22

\[\int\limits_{}^{}\frac{\sec^2(u)}{\tan^2(u) \sqrt{\sec^2(u)}} du\]

Answer 23

\[\int\limits_{}^{}\frac{\sec^2(u)}{\tan^2(u) \sec(u)}\]

Answer 24

\[\int\limits_{}^{}\frac{\sec(u)}{\tan^2(u)} du\]

Answer 25

alright I 'm following and not lost thus far

Answer 26

\[\int\limits_{}^{}\sec(u) \cdot \frac{1}{\tan^2(u)} du=\int\limits_{}^{}\frac{1}{\cos(u)} \cdot \frac{\cos^2(u)}{\sin^2(u)} du\]

Answer 27

\[\int\limits_{}^{}\frac{\cos(u)}{\sin^2(u)} du\]

Answer 28

guess what?
we are almost there

Answer 29

do you see what to do next

Answer 30

yes

Answer 31

let theta = sin u

Answer 32

ok yes we can call the new substitution theta ! :)

Answer 33

dtheta = cos u du

Answer 34

incredible :)

Answer 35

cool then it would be intergral 1/theta^2

Answer 36

yes thats what we want to integrate

Answer 37

oh you said integral gj

Answer 38

:-)

Answer 39

\[\frac{\theta^{-2+1}}{-2+1}+C=\frac{\theta^{-1}}{-1}+C=\frac{-1}{\theta}+C\]

Answer 40

but what was theta?

Answer 41

sin(u) right?

Answer 42

sin u

Answer 43

yes

Answer 44

omg you are so great

Answer 45

\[\frac{-1}{\sin(u)}+C\]

Answer 46

lol why thanks lol

Answer 47

last thing to do is rewrite in terms of x

Answer 48

ok

Answer 49

so remember we let x=tan(u)
you might feel a need to draw a triangle
thats what i would do